Applications — biological (haemoglobin, chlorophyll, vit B₁₂), medicinal (cisplatin), industrial (catalysts): Worked examples

This page builds on the parent Applications topic and pulls tools from Oxidation states of transition metals, Isomerism in coordination compounds, Square planar complexes, Crystal Field Theory, and Stability and chelate effect.

The scenario matrix

Before we solve anything, let us list every class of case this topic can throw at you. Think of it as a grid: each row is a "kind of thinking" the question forces, and the last column names the worked example that covers it.

Cell	Case class	What makes it tricky	Covered by
A	Assign oxidation state to the central metal	ligand charges must be counted with correct sign	Ex 1
B	Geometric-isomer decides function (cis vs trans)	same formula, opposite biology	Ex 2
C	"Degenerate" metal — right metal, wrong oxidation state	Fe(II)→Fe(III) kills the function	Ex 3
D	Competition / equilibrium limit (CO vs O₂)	binding strength ratio, not destruction	Ex 4
E	Colour = complementary wavelength (limiting/optical)	reflected ≠ absorbed; frequency–energy link	Ex 5
F	Metal chosen for what it cannot do (redox-inert Mg)	absence of a property is the answer	Ex 6
G	Count donor atoms / denticity of a macrocycle	axial vs equatorial sites	Ex 7
H	Real-world word problem (dose / competition numbers)	plug real numbers, keep units	Ex 8
I	Exam twist — catalyst cycle bookkeeping	oxidation state changes around a loop	Ex 9

Every cell A–I is filled below. If you can do all nine, you can do any exam question on this topic.

Cell A — Assigning oxidation state

Worked example Example 1 — Find the oxidation state of Co in vitamin B₁₂ (cyanocobalamin), treated as the neutral molecule

$[Co(corrin) (CN) (base)]$ . Forecast: guess before reading — is Co here +1, +2, or +3?

Step 1. Let $x$ stand for the number we want: $x =$ the oxidation state of the cobalt atom (a bookkeeping charge we assign to the metal). Write the balance rule: $x + (sum of all ligand charges) = (overall charge on the molecule) .$ Why this step? Oxidation state is just charge bookkeeping; the only honest way to get $x$ is to balance charges — see Oxidation states of transition metals.

Step 2. Assign each ligand its charge using the standard convention:

the corrin macrocycle (like porphyrin, with two pyrrole-type N deprotonated) is counted as a dianion, charge $- 2$ ;
the cyanide $CN^{-}$ is $- 1$ ;
the neutral nitrogen base contributes $0$ .

Why this step? Anionic ligands must be counted with their real sign, or the metal number comes out wrong. Getting the corrin charge right ( $- 2$ , not $- 1$ ) is the whole trick.

Step 3. The molecule as written is neutral (overall charge $= 0$ ). Substitute the charges from Step 2 into the balance rule and solve for $x$ : $x + (- 2) + (- 1) + 0 = 0 ⟹ x - 3 = 0 ⟹ x = + 3.$ Why this step? This is the single, consistent equation — no guessing between values. It gives Co(III), exactly the parent recall table.

Verify: put $x = + 3$ back: $(+ 3) + (- 2) + (- 1) + 0 = 0$ ✓ neutral molecule, matches the parent table (Co, +3, corrin).

Cell B — Isomer decides function

Worked example Example 2 — Which isomer of

$[Pt (NH_{3})_{2} Cl_{2}]$ can crosslink two adjacent DNA bases, and quantify the geometry? Forecast: cis or trans — and roughly how far apart do the two Cl sit?

Step 1. Geometry: Pt(II) is $d^{8}$ (eight $d$ -electrons, as defined at the top of the page) → square planar (four sites at $9 0^{\circ}$ steps). See Square planar complexes. Why this step? The angular positions of the leaving groups are set by the geometry; without it we cannot talk about $9 0^{\circ}$ vs $18 0^{\circ}$ .

Step 2. In cis, the two Cl occupy adjacent corners → angle between the two Pt–Cl bonds $= 9 0^{\circ}$ . In trans, they are opposite → $18 0^{\circ}$ . Look at the figure: the highlighted magenta arc marks the $9 0^{\circ}$ bite of the cis isomer, and the violet dashed line shows the straight $18 0^{\circ}$ of the trans.

Figure 1 — Cell B. Left: the cis isomer (two magenta Cl on adjacent corners) — the shaded magenta wedge highlights the $9 0^{\circ}$ angle and the magenta double-arrow marks the short Cl···Cl distance $2.83 \overset{˚}{A}$ that can reach two neighbouring DNA bases. Right: the trans isomer (Cl opposite) — the violet dashed double-arrow shows the wide $18 0^{\circ}$ , $4.0 \overset{˚}{A}$ span whose arms point apart and cannot bridge. The violet dot is the Pt centre; orange dots are the two ammine ligands.

Why this step? DNA crosslinking needs two leaving groups pointing at two neighbouring guanine N7 atoms; only the $9 0^{\circ}$ pair can reach both.

Step 3. Model the reach. Take the Pt–ligand bond length as a fixed value $b$ and compute the straight-line Cl···Cl distance by the law of cosines: $d = b^{2} + b^{2} - 2 b b cos θ .$ We use $b = 2.0 \overset{˚}{A}$ as a round working figure (the true Pt–Cl bond is closer to $2.3 \overset{˚}{A}$ ; the exact value does not change which isomer wins, since $d$ scales linearly with $b$ — only the ratio of cis to trans matters here). Why this step? This turns "adjacent vs opposite" into a comparable number; because both isomers use the same $b$ , any reasonable bond length gives the same conclusion.

Step 4. For cis ( $θ = 9 0^{\circ}$ ): $d = 8 \approx 2.83 \overset{˚}{A}$ . For trans ( $θ = 18 0^{\circ}$ ): $d = 4.0 \overset{˚}{A}$ . Why this step? The cis distance is short enough to span two adjacent bases; the trans distance is larger and the arms point apart → cannot bridge.

Verify: $cos 9 0^{\circ} = 0 \Rightarrow d = 8 = 2.828 \overset{˚}{A}$ ; $cos 18 0^{\circ} = - 1 \Rightarrow d = 16 = 4.0 \overset{˚}{A}$ . With any $b$ the trans distance is always the larger ( $2 b$ vs $b 2$ ) — cis wins. Only cis is the drug — consistent with Isomerism in coordination compounds.

Cell C — Right metal, wrong oxidation state (degenerate case)

Worked example Example 3 — Methaemoglobin: why does oxidising Fe(II) to Fe(III) destroy O₂ transport?

Forecast: does the iron leave, or does it change?

Step 1. Working haemoglobin (Hb, defined at the top) has Fe in the $+ 2$ state. Confirm by charge count with $x =$ iron oxidation state: porphyrin is $- 2$ , the histidine N is neutral, and the resting site holds nothing net-charged, so a neutral haem needs $x - 2 = 0 \Rightarrow x = + 2$ . Why this step? We must first prove the normal state is $+ 2$ before we can say what oxidising it changes.

Step 2. Oxidation removes one electron: $Fe^{2 +} \to Fe^{3 +} + e^{-}$ . Why this step? "Met" means the metal was oxidised — this is a redox change, not a broken cage.

Step 3. Fe(III) is harder, more positive, and binds water/hydroxide tightly at the 6th site → that site is blocked → no reversible O₂ uptake. Why this step? Reversibility is the whole point; a tightly held H₂O is not released, so no O₂ cycling.

Verify: charge check on met-Hb site: $Fe^{3 +}$ + porphyrin $^{2 -}$ + $OH^{-}$ gives $+ 3 - 2 - 1 = 0$ , a stable neutral non-carrier ✓. Metal is the same element, just $+ 3$ — the same conclusion reached later in Example 7 of this page (Cell G), where the free O₂ site is the very site now blocked.

Cell D — Competition / equilibrium limit

Worked example Example 4 — CO binds Hb about

$200 \times$ tighter than O₂. If lungs have $p_{O_{2}} = 100$ and $p_{C O} = 0.5$ (same pressure units), what is the ratio of CO-bound to O₂-bound haemoglobin? Forecast: with only 0.5 units of CO vs 100 of O₂, will CO still capture a big fraction?

Step 1. For two gases competing at the same single site, divide their binding expressions; the free $[Hb]$ cancels and we get $\frac{[ HbCO ]}{[ HbO _{2} ]} = \frac{K _{C O}}{K _{O_{2}}} \cdot \frac{p _{C O}}{p _{O_{2}}} .$ Here $K_{C O}$ and $K_{O_{2}}$ are the association constants defined above. Why this step? Both gases fight for the same 6th site, so the occupancy ratio is (relative affinity) × (relative pressure). This is competition, not destruction — the parent's key correction.

Step 2. Plug in $K_{C O} / K_{O_{2}} = 200$ and $p_{C O} / p_{O_{2}} = 0.5/100$ : $\frac{[ HbCO ]}{[ HbO _{2} ]} = 200 \times \frac{0.5}{100} = 200 \times 0.005 = 1.0.$ Why this step? It shows a tiny trace of CO already ties up as much Hb as all the oxygen — why CO is deadly.

Step 3. Interpret: ratio $= 1$ means half the carrying capacity is stolen even at $\frac{1}{200}$ the pressure of oxygen. Why this step? Turns the affinity number into a health consequence.

Step 4 (edge case — cooperativity). Real haemoglobin has four binding sites that talk to each other: binding one O₂ makes the next easier (positive cooperativity, described by the sigmoidal Hill curve). Our one-site formula is a simplification. Cooperativity makes CO poisoning worse: CO on one subunit raises the O₂ affinity of the others, so they hold onto O₂ in the tissues and fail to release it — a second, subtler harm beyond simple competition. Why this step? Exams love the twist "why is CO worse than pure competition predicts?" — the answer is the Hill/cooperative effect, an edge case the simple model misses.

Verify: $200 \times 0.005 = 1.0$ ✓; treatment = raise $p_{O_{2}}$ (pure oxygen) to push the ratio back below 1, exactly the parent's stated cure.

Cell E — Colour is the complement of what is absorbed

Worked example Example 5 — Chlorophyll strongly absorbs light near

$430 nm$ (blue) and $662 nm$ (red). Estimate the energy of a $662 nm$ photon (in joules and in eV) and confirm chlorophyll therefore looks green. Forecast: which colour is left over to reach your eye?

Step 1. Photon energy: $E = \frac{h c}{λ}$ with $h = 6.626 \times 1 0^{- 34} J\cdots$ , $c = 3.0 \times 1 0^{8} m/s$ . Why this tool and not another? The gap between the ring's energy levels sets which wavelength is swallowed; $E = h c / λ$ is the bridge between "energy gap" and "colour absorbed" — the Crystal Field Theory colour logic.

Step 2. $λ = 662 nm = 6.62 \times 1 0^{- 7} m$ , so $E = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3.0 \times 1 0 ^{8} )}{6.62 \times 1 0 ^{- 7}} \approx 3.0 \times 1 0^{- 19} J .$ Why this step? Confirms these are ordinary visible photons.

Step 3. Convert to electronvolts using the definition $1 eV = 1.602 \times 1 0^{- 19} J$ : $E = \frac{3.0 \times 1 0 ^{- 19} J}{1.602 \times 1 0 ^{- 19} J/eV} \approx 1.87 eV .$ Why this step? Chemists quote visible-light energies in eV; ~1.9 eV confirms this sits squarely in the visible band.

Step 4. Absorbed = blue + red ⇒ the un-absorbed middle band = green (~500–560 nm) is reflected. By the complementary-colour rule from the callout above, the leaf therefore looks green. Why this step? The colour you see is the complement of what is absorbed — turning the two absorbed peaks into the observed green.

Figure 2 — Cell E. Absorption spectrum of chlorophyll (violet curve): two strong peaks at ~430 nm (blue) and ~662 nm (red), each labelled with a downward arrow at the swallowed wavelength. The shaded green window (500–560 nm) is not absorbed, so it is reflected/transmitted — which is why leaves look green. The magenta arrow marks the 662 nm red photon whose energy we computed ( $3.0 \times 1 0^{- 19}$ J ≈ 1.87 eV).

Verify: $E = 6.626 \times 1 0^{- 34} \times 3.0 \times 1 0^{8} /6.62 \times 1 0^{- 7} = 3.00 \times 1 0^{- 19} J$ ✓; $3.00 \times 1 0^{- 19} /1.602 \times 1 0^{- 19} = 1.87 eV$ ✓ (a visible-light photon). Green survives ✓.

Cell F — Metal chosen for what it cannot do

Worked example Example 6 — Why Mg(II) in chlorophyll and not Fe(II)? Show it is the

absence of easy redox that matters. Forecast: is Mg picked because it is a better electron-handler than Fe, or a worse one?

Step 1. List available oxidation states. Fe: commonly $+ 2$ and $+ 3$ (easy switch). Mg: only $+ 2$ (a full-shell, redox-inert ion). Why this step? The job of chlorophyll's metal is not to grab O₂ or cycle electrons on itself — so redox activity would be a liability.

Step 2. Mg²⁺ holds the conjugated ring rigid and planar; the excited electron travels through the ring, not through the metal. Why this step? A redox-active Fe would quench the excited state by accepting the electron itself.

Step 3. Conclusion: Mg is chosen because it stays $+ 2$ — the useful property is a non-property. Why this step? This is the "degenerate/limiting" reasoning: sometimes the answer is "it does nothing, and that's the point."

Verify: charge check — neutral chlorophyll core: $Mg^{2 +}$ + porphyrin $^{2 -}$ gives $+ 2 - 2 = 0$ ✓, and Mg has no accessible $+ 3$ , matching the parent's "redox-inert" claim.

Cell G — Counting donor atoms / free sites

Worked example Example 7 — In haemoglobin, how many coordination sites does Fe use, how many come from the porphyrin, and how many are left for chemistry?

Forecast: guess the total coordination number of the iron.

Step 1. Fe(II) here is 6-coordinate (octahedral): 4 equatorial + 2 axial. Why this step? We must fix the geometry before counting who fills each slot.

Step 2. Porphyrin is a macrocycle donating through 4 N in the equatorial plane. That uses $4$ of the 6 sites. Why this step? The macrocycle's denticity (4) directly reduces the free-site count — the chelate/macrocycle idea from Stability and chelate effect.

Step 3. Free axial sites $= 6 - 4 = 2$ . One axial site is anchored by the protein's histidine; the last one binds O₂ reversibly. Why this step? This "6th site" is the whole functional story of O₂ transport — and the very site that Cell C's Fe(III) blocks.

Figure 3 — Cell G. Octahedral Fe(II) (magenta centre) with its six coordination sites: four equatorial nitrogen donors from the flat porphyrin ring (violet), one axial site anchored below by the protein's histidine (navy), and the sixth axial site above free to bind O₂ reversibly (orange). Total coordination number $= 4 + 1 + 1 = 6$ .

Verify: $4 (porphyrin N) + 1 (histidine) + 1 (O_{2}) = 6$ ✓ octahedral, exactly one O₂-binding site.

Cell H — Real-world word problem

Worked example Example 8 — A patient needs a cisplatin dose of

$75 mg/m^{2}$ and has a body-surface area of $1.8 m^{2}$ . How many millimoles of Pt does the dose contain? ( $M (cisplatin) = 300 g/mol$ .) Forecast: roughly how many mg total, and will the mmol be under 1?

Step 1. Total mass $= 75 mg/m^{2} \times 1.8 m^{2} = 135 mg$ . Why this step? Dose per surface area × surface area = total drug; units cancel to mg.

Step 2. Convert to moles: $mol = \frac{mass (g)}{M} = \frac{0.135 g}{300 g/mol} = 4.5 \times 1 0^{- 4} mol$ . Why this step? Biological activity is per molecule (per Pt centre), so we need moles, not mass.

Step 3. In mmol: $4.5 \times 1 0^{- 4} mol \times 1000 = 0.45 mmol$ . Each molecule has exactly one Pt, so this is also $0.45 mmol$ of Pt. Why this step? One complex = one metal centre, so Pt moles = complex moles.

Verify: $135 mg = 0.135 g$ ; $0.135/300 = 0.00045 mol = 0.45 mmol$ ✓, under 1 mmol as forecast.

Cell I — Exam twist: catalyst-cycle bookkeeping

Worked example Example 9 — In Wacker oxidation, Pd cycles between two oxidation states as it turns ethene into ethanal, then

$Cu^{2 +}$ re-oxidises it. If Pd goes from $+ 2$ (active) to $0$ (spent), how many electrons must Cu²⁺ supply per turnover, and what does Cu become? Forecast: how many electrons, and does Cu go up or down?

Step 1. Pd change: $Pd^{2 +} \to Pd^{0}$ is a gain of $2$ electrons (reduction). Why this step? The oxidation-state drop of 2 tells us exactly how many electrons the catalyst absorbed while doing the chemistry.

Step 2. To regenerate the active catalyst, those $2$ electrons must be removed again — $Cu^{2 +}$ takes them: each $Cu^{2 +} \to Cu^{+}$ accepts $1 e^{-}$ , so two Cu²⁺ ions are needed per Pd. Why this step? A catalyst must return to its start state; the co-oxidant Cu closes the loop.

Step 3. Track what Cu becomes and close the cycle. The two $Cu^{2 +}$ are reduced to two $Cu^{+}$ . These $Cu^{+}$ are then re-oxidised back to $Cu^{2 +}$ by molecular $O_{2}$ — so copper, like palladium, returns to its starting state and the only thing truly consumed by the overall process is oxygen. Why this step? It proves both metals are genuine catalysts (they return to start) and identifies O₂ as the real electron sink, which is the point of the Wacker process.

Verify: electrons: Pd needs $∣0 - (+ 2) ∣ = 2$ e⁻; $2 \times (Cu^{2 +} \to Cu^{+})$ supplies $2 \times 1 = 2$ e⁻ ✓ balanced. Both Pd and Cu return to their starting oxidation states ⇒ genuinely catalytic, with O₂ consumed.

Recall Cell coverage self-check

Nine cells A–I, one example each — did every case get a worked answer? A oxidation-state count ::: Ex 1 (Co = +3) B isomer decides function ::: Ex 2 (cis, $d = 2.83$ Å) C right metal wrong state ::: Ex 3 (Fe³⁺ blocks site) D competition limit ::: Ex 4 (ratio = 1.0, plus cooperativity) E colour = complement ::: Ex 5 ( $3.0 \times 1 0^{- 19}$ J ≈ 1.87 eV, green) F metal for a non-property ::: Ex 6 (Mg redox-inert) G counting free sites ::: Ex 7 (2 axial, one for O₂) H word problem ::: Ex 8 (0.45 mmol Pt) I catalyst bookkeeping ::: Ex 9 (2 electrons, 2 Cu²⁺)

Connections

Isomerism in coordination compounds — cis vs trans in Ex 2
Square planar complexes — $d^{8}$ Pt(II) geometry in Ex 2
Oxidation states of transition metals — charge counting in Ex 1, 3, 9
Crystal Field Theory — energy gap ↔ colour in Ex 5
Stability and chelate effect — macrocycle denticity in Ex 7